## Analysis of Hartree-Fock equations, Koopman's theorem

With similar arguments as in atomic physics, we can now use Hartree-Fock theory to make a link between nuclear forces and separation energies. Changing to nuclear system, we define $$E[\Phi^{\mathrm{HF}}(A)] = \sum_{i=1}^A \langle i | \hat{h}_0 | i \rangle + \frac{1}{2}\sum_{ij=1}^A\langle ij|\hat{v}|ij\rangle_{AS},$$ where $$\Phi^{\mathrm{HF}}(A)$$ is the new Slater determinant defined by the new basis of Eq. (13) for $$A$$ nucleons, where $$A=N+Z$$, with $$N$$ now being the number of neutrons and $$Z$$ th enumber of protons. If we assume again that the single-particle wave functions in the new basis do not change from a nucleus with $$A$$ nucleons to a nucleus with $$A-1$$ nucleons, we can then define the corresponding energy for the $$A-1$$ systems as $$E[\Phi^{\mathrm{HF}}(A-1)] = \sum_{i=1; i\ne k}^A \langle i | \hat{h}_0 | i \rangle + \frac{1}{2}\sum_{ij=1;i,j\ne k}^A\langle ij|\hat{v}|ij\rangle_{AS},$$ where we have removed a single-particle state $$k\le F$$, that is a state below the Fermi level.